What is the value of the following logarithm? $\log_{4} \left(\dfrac{1}{64}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $4^{y} = \dfrac{1}{64}$ In this case, $4^{-3} = \dfrac{1}{64}$, so $\log_{4} \left(\dfrac{1}{64}\right) = -3$.